Wilbur Wright to Octave Chanute

Dayton, December 3, 1900

Yours 2nd rec'd. The form of designation used in referring to us is
immaterial but in so far as we have any preference it is for "Messrs.
Wilbur and Orville Wright."

I think one half would be better than one third in stating the reduction in
resistance of our machine and operator's position as compared with your machine
and upright position. Our figures are only one third but I think that we have
underestimated our resistance or you have slightly overestimated yours, and
that the real reduction is only to one half instead of to one third the
head resistance of your trials.

Our reasons for believing that the equivalent sq. ft. of the operator's body
in the horizontal position is only 1/2 sq. ft. are as follows.

1. We tested the resistance of the machine with log chains upon it; also
with a boy of equal weight and though the log chains offered almost no cross
section we found the results almost the same. The resistance of the boy could
not have exceeded a pound.

2. A bicycle rider has ridden 220 yards, without pace, in 12 seconds. This
is 3,300 ft. per minute and 37 1/2 miles per hour. The resistance at 37 1/2
miles is 7 lbs. per sq. ft. so that each sq. foot requires 7 X 3,300 = 23,100
ft. pounds per minute. At 80 gear the rider makes
520 pedal strokes per mile, or 325 strokes per minute. The length of stroke is
1 1/6 ft. and the weight of rider 150 lbs. The extreme limit of power is 150 X
1 1/6 X 325 = 56,875 ft. pounds or nearly two horsepower. This is undoubtedly
more than the man actually exerts. The friction of tires, bearings and chain
requires a pull of 3 lbs. to overcome, or 3 X 3,300 = 9,900 ft. pounds per
minute. Subtracting this from 56,875 leaves 46,875 [46,975] as the extreme
power available for overcoming wind resistance. But the bicycle offers a cross
section of 2 1/4 sq. ft. whose coefficient is probably 1/5, so its equivalent
area is about 1/2 sq. ft. and the required ft. pounds 23,100 ÷ 2 = 11,550.
But 46,875 - 11,550 = 35,325 is all that is now available for the resistance of
the man; and his equivalent area is 35,325 ÷ 23,100 = 1 1/3 [1 1/2] sq.
ft. But the cross section of a man, in racing position is a little over 3 sq.
ft. so that his coefficient of resistance is less than one half. If no
reduction be made in the area exposed it would require nearly four horsepower
to drive him through the air at 37 1/2 miles per hour. If the coefficient for
the bicycle racing position be less than 1/2 I see no reason for estimating the
equivalent area of a man in the horizontal position at more than 1/2 sq. ft.,
and though I may be mistaken I doubt whether the resistance in the upright
position is more than equivalent to 2 1/2 sq. ft.